Tag: inequalities in triangles

Questions Related to inequalities in triangles

If $z$ is a complex number satisfying the equation $\left| z+i \right| +\left| z-i \right| =8$, on the complex plane then maximum value of $\left| z \right| $ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $2$

  2. $4$

  3. $6$

  4. $8$

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Correct Option: B

Suppose z and $\omega$ are two complex numbers such that $|z| \leq 1, |\omega| \leq 1$, and $|z+i\omega|=|z-i\omega|=2$.

Which of the following is true about $|z|$ and $|\omega|$?

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $|z|=|\omega|=\frac {1}{2}$

  2. $|z|=\frac {1}{2}, |\omega|=\frac {3}{4}$

  3. $|z|=|\omega|=\frac {3}{4}$

  4. $|z|=|\omega|=1$

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Correct Option: D
Explanation:

We have,


$|z+i\omega|\le |z|+|i\omega|\le |z|+|i||\omega|\le 2$

$|z-i\omega|\le |z|+|-i\omega|\le |z|+|i||-\omega|=|z|+|i||\omega|\le 2$

Also,

$|z+i\omega|=|z-i\omega|$

and

$ |z+i\omega|=2$ 

Hence, 

$|z|  = |\omega| = 1 $

$\begin{array} { l } { \text { If } z _ { 1 } \text { and } z _ { 2 } \text { are complex numbers, then } \left| z _ { 1 } + z _ { 2 } \right| ^ { 2 } = \left| z _ { 1 } \right| ^ { 2 } + \left| z _ { 2 } \right| ^ { 2 } \text { if and only if } z _ { 1 } \overline { z } _ { 2 } \text { is } } \ { \text { purely imaginary. } } \end{array}$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. True

  2. False

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Correct Option: A

If $|z| < \sqrt 2-1$, then $|z^2+2z cos\alpha|$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. Less than 1

  2. $\sqrt 2+1$

  3. $\sqrt 2-1$

  4. None of these

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Correct Option: A
Explanation:

$|z^2+2z cs \alpha| \leq |z^2|+|2z cos \alpha|$
$=|z^2|+|2z^2| |cos\alpha|$
$\leq |z|^2 + 2 |z| $
$ < (\sqrt 2-1)^2 + 2(\sqrt 2-1) =1$

If $P$ and $Q$ are represented by complex numbers $z _{1}$ and $z _{2}$ such that $\left| \dfrac { 1 }{ { z } _{ 1 } } +\dfrac { 1 }{ { z } _{ 2 } }  \right| =\left| \dfrac { 1 }{ { z } _{ 1 } } -\dfrac { 1 }{ { z } _{ 2 } }  \right| $ then the circumference of $\triangleOPQ(O is origin)$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\dfrac{{ z } _{ 1 } -{ z } _{ 2 } }{2}$

  2. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{2}$

  3. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{3}$

  4. ${ z } _{ 1 } +{ z } _{ 2 } $

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Correct Option: A

The sum of all sides of a quadrilateral is lessthan the sum of its diagonals.

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. True

  2. False

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Correct Option: B

If $\left| {z - 1} \right| + \left| {z + 3} \right| \le 8$ then the range of values of $\left| {z - 4} \right|$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $[1,\,7]$

  2. $[1,\,8]$

  3. $[1,\,9]$

  4. $[2,\,5]$

Show answer
Correct Option: A
Explanation:

$|z-1|+|z+3|\le 8$

Using the triangle inequality 
$|z _1\pm z _2|\le |z _1|+|z _2|$
We have $|z-1|+|z+3|\le 8$
$\implies |z-1+z+3|\le 8$
$\implies |z+1|\le 4$
Using triangle inequality again 
$|z|+1\le 4\implies |z|\le 3$
So, the maximum value of $|z _1+ _2|$ is $|z _1|+|z _2|$
And  the minimum value of $|z _1+ _2|$ is $|z _1|-|z _2|$
Hence the maximum value of $|z-4|$ is $|z|+|4|=3+4=7$
 the minimum value of $|z-4|$ is $|z|-|4|=3-4=-1$
Hence the range of $|z-4|$
$1\le|z-4|\le 7$

If $\left|z\right| <\sqrt{2} -1$, then $\left|z^2 + 2 z  cos  \alpha \right|$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. less than 1

  2. $\sqrt{2} + 1$

  3. $\sqrt{2} -1$

  4. none of these

Show answer
Correct Option: A
Explanation:

$\left| z \right| <\sqrt { 2 } -1\ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le \left| { z }^{ 2 } \right|+ \left| 2z\cos { \alpha  }  \right| \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| { z } _{ 1 }+{ z } _{ 2 } \right| \le \left| { z } _{ 1 } \right| +\left| { z } _{ 2 } \right|  \right} \ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le |z|(|z|+2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| \cos { \alpha  }  \right| \le 1\quad  \right} \ \Rightarrow \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <(\sqrt{2}-1){ \left( \sqrt { 2 } +1 \right)  }<1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| z \right| <\sqrt { 2 } -1\quad  \right} \ \therefore \quad \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <1\ $
Hence, option 'A' is correct.

If z be a complex number for which $|2z  cos  \theta + z^2| = 1$, then the minimum value of |z|
 is ......................

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{2} -1$

  4. $\sqrt{2} +1$

Show answer
Correct Option: C
Explanation:

$|z^{2}+2zcos\theta|$
$=|z(z+2cos\theta)|$
$=|z|.|z+2cos\theta|$
$=1$
Now 
$|z|=1$ and 
$|z+2cos\theta|=1$
Now 
$|z+2cos\theta|\leq |z|+|2cos\theta|$
Considering 
$|z+2\cos\theta|=|z|+|2cos\theta|=1$
Hence
$|z|=|2cos\theta|\pm1$
Considering $z=|2cos\theta|-1$ we get the minimum value at multiples of $\theta=45^{0}$
Hence
$z=\sqrt{2}-1$.

$sin^{-1}\left { \frac{1}{i} (z-1)\right }$ ,Where Z is non - real, can be the angle  of a triangle, if 

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $Re(z)=1, Im(z)=2$

  2. $Re(z)=1,-1\leq Im(z)\leq 1$

  3. $Re(z)=1,Im(z)=0$

  4. $Re(z)=1,Im(z)=-2$

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Correct Option: A